$\begingroup$@MyGlasses Actually it is $a^2$. I left it there to note it's positive.$\endgroup$
– Adel BibiJul 2 '17 at 12:36

$\begingroup$what have you tried? for the indefinite case seems difficult, for the case where thie interval of integration is $(-1,1)$ there might be hope$\endgroup$
– tiredJul 2 '17 at 12:51

1

$\begingroup$@tired I have added the edits. Could you highlight a way to approach the problem for the range of (-1,1)?$\endgroup$
– Adel BibiJul 2 '17 at 13:05

2

$\begingroup$i am a bit busy today to really dig into this question but i would advise to explore the following link: math.stackexchange.com/questions/1148493/… which handles a very similar looking integral by inverse LT. There is also this question of mine, which highlights another technique which might be valuable in this kind of situations namley trigoometric substitions: math.stackexchange.com/questions/1958258/…$\endgroup$
– tiredJul 2 '17 at 13:16

2

$\begingroup$no result in the known elementary functions found$\endgroup$
– Dr. Sonnhard GraubnerJul 2 '17 at 14:10

1 Answer
1

This is not a complete answer to this question however I believe it can bring some insight to this problem. Let us compute the definite integral rather than the antiderivative. Therefore the objective is to compute:
\begin{eqnarray}
{\mathcal I}(a,b):=\int\limits_0^1 \exp\left( -\frac{1}{2} \frac{a+b t}{1-t^2}\right) \frac{1}{\sqrt{1-t^2}} dt
\end{eqnarray}
Firstly let us handle the case $b=0$. In here we substitute for $u=\sqrt{a/2 \cdot t^2/(1-t^2)}$. We have:
\begin{eqnarray}
{\mathcal I}(a,0)&=&\sqrt{2 a} \exp\left(-\frac{a}{2}\right) \frac{1}{4} \int\limits_{-\infty}^\infty \frac{e^{-u^2}}{u^2+a/2} du \\
&=&\sqrt{2 a} \exp\left(-\frac{a}{2}\right) \frac{1}{4} \int\limits_{-\infty}^\infty \frac{e^{-\frac{k^2}{4}}}{\sqrt{2}} \cdot \frac{\sqrt{\pi}}{\sqrt{a}} e^{-\frac{\sqrt{a} |k|}{\sqrt{2}}} dk\\
&=& \frac{\pi}{2} Erfc[\frac{\sqrt{a}}{\sqrt{2}}]
\end{eqnarray}
where in the second line we used the Parseval's identity and in the last line we completed to square and used Gaussian integration.

Now we are going to compute the whole quantity ${\mathcal I}(a,b)$ by expanding the exponential in a series. So we have:
\begin{eqnarray}
{\mathcal I}(a,b)-{\mathcal I}(a,0)&=&\sum\limits_{n=1}^\infty \frac{(-b/2)^n}{n!} \int\limits_0^1
\exp\left( -\frac{1}{2} \frac{a}{1-t^2}\right) \frac{1}{\sqrt{1-t^2}} \left(\frac{t}{1-t^2}\right)^n dt\\
&=&
\sum\limits_{n=1}^\infty \frac{(-b/2)^{2 n}}{(2 n)!} \int\limits_0^1
\exp\left( -\frac{1}{2} \frac{a}{1-t^2}\right) \frac{1}{\sqrt{1-t^2}} \left(\frac{t}{1-t^2}\right)^{2 n} dt+\\
&&\sum\limits_{n=0}^\infty \frac{(-b/2)^{2 n+1}}{(2 n+1)!} \int\limits_0^1
\exp\left( -\frac{1}{2} \frac{a}{1-t^2}\right) \frac{1}{\sqrt{1-t^2}} \left(\frac{t}{1-t^2}\right)^{2 n+1} dt\\
&=&
\sum\limits_{n=1}^\infty \frac{(-b/2)^{2 n}}{(2 n)!} 2^{2 n-1/2} \frac{\sqrt{a} }{a^{2 n}} \exp(-a/2) \sum\limits_{p=0}^{n-1} \binom{n-1}{p} (a/2)^{n-p-1} \frac{1}{2} \Gamma(\frac{1}{2}+n+p) +\\
&&\sum\limits_{n=0}^\infty \frac{(-b/2)^{2n+1}}{(2n+1)!} 2^{2 n+1/2} \frac{\sqrt{a}}{a^{2 n+1}} \sum\limits_{p=0}^n \binom{n}{p} (-a/2)^{n-p} \frac{1}{2} \Gamma(\frac{1}{2}+n+p,\frac{a}{2})
\end{eqnarray}
where the last two lines were derived by substituting for $u$ as on the very top of this answer.
Now we need to check convergence . As it turns out for the series to converge $a$ has to be positive and large enough. Below I plot the integral in question both analytically (from the series expansion) and numerically (from definition).

Update: Let us now compute the antiderivative. We define:
\begin{equation}
{\mathcal I}_{(a,b)}(x):= \int\limits_0^x \frac{1}{\sqrt{1-t^2}} \exp\left(-\frac{1}{2} \frac{a+b t}{1-t^2} \right) dt
\end{equation}
where $x \in(0,1]$.
Now, by doing the same calculations as before with no modification whatsoever we get the following results. Firstly we have:
\begin{equation}
{\mathcal I}_{(a,0)}(x):= 2 \pi T(\sqrt{a}, \frac{x}{\sqrt{1-x^2}})
\end{equation}
where $T$ is the Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function .
Then we have:
\begin{eqnarray}
&&\int\limits_0^x \frac{1}{\sqrt{1-t^2}} \exp\left( -\frac{1}{2} \frac{a+b t}{1-t^2} \right) \cdot \left( \frac{t}{1-t^2} \right)^{2 n} dt = \\
&&2^{-1/2+n} a^{1/2-2 n} \sum\limits_{p=0}^{n-1} \binom{n-1}{p} a^{n-1-p} 2^p \exp(-a/2) \Gamma\left(\frac{1}{2} + n+p,0, \frac{a}{2} \frac{x^2}{1-x^2} \right)\\
&&\int\limits_0^x \frac{1}{\sqrt{1-t^2}} \exp\left( -\frac{1}{2} \frac{a+b t}{1-t^2} \right) \cdot \left( \frac{t}{1-t^2} \right)^{2 n+1} dt = \\
&&2^{-1/2+n} a^{-1/2-2 n} \sum\limits_{p=0}^n \binom{n}{p} (-a)^{n-p} 2^p \Gamma\left(\frac{1}{2} +n+p,\frac{a}{2},\frac{a}{2} \frac{1}{1-x^2} \right)
\end{eqnarray}
We can now sum over $n$ to obtain the whole result. It turns out that the series converge quite rapidly. Below I present the contour plots of the quantity in question being computed for $x=0.1,0.3,0.5,0.7,0.9$ from the bottom to the top respectively. As before on the left I produce the analitical result (the series expansion in $n$ being truncated at $n\le 20$) and on the right I show the numerical result (the integral being computed numerically).

$\begingroup$Thanks Przemo. I found a closed formula for this in terms of Hermite polynomials and incomplete gamma function. I will be posting the answer later. I will make sure to tag you.$\endgroup$
– Adel BibiFeb 28 at 14:13